Equations and Inequalities

Solution: p = 3

Topic 1 Answer Key • Savvas Algebra 1 Page 1 of 3

III

Part 3: Solving Multi-Step Linear Inequalities (Lessons 1-4 & 1-5)

Answer Key

#10 3x + 1 < -5

Subtract 1: \(3x < -5 - 1 \implies 3x < -6\)

Divide by 3: \(x < \frac{-6}{3} \implies x < -2\)

Interval Notation: \((-\infty, -2)\)

Solution: x < -2

#11 \( \frac{7}{9}x - 4 < \frac{3}{4} \)

Add 4 (\(\frac{16}{4}\)): \(\frac{7}{9}x < \frac{3}{4} + \frac{16}{4} \implies \frac{7}{9}x < \frac{19}{4}\)

Multiply by reciprocal \(\frac{9}{7}\): \(x < \frac{19}{4} \times \frac{9}{7} = \frac{171}{28}\)

\(\frac{171}{28} \approx 6.107\) (or \(6\frac{3}{28}\))

Solution: x < 171/28   (or 6 3/28)

#12 4(m - 6) > 12

Method 1 (Divide by 4): \(m - 6 > 3 \implies m > 9\)

Method 2 (Distribute): \(4m - 24 > 12 \implies 4m > 36 \implies m > 9\)

Interval Notation: \((9, \infty)\)

Solution: m > 9

#16 3(x - 5) ≥ 3x - 25

Distribute 3: \(3x - 15 \ge 3x - 25\)

Subtract \(3x\): \(-15 \ge -25\)

Since \(-15 \ge -25\) is a TRUE statement for all \(x\), the solution is all real numbers.

Correct Option: C. All real numbers

#13 r + 6 - 2(r - 22) > 0

Distribute \(-2\): \(r + 6 - 2r + 44 > 0\)

Combine like terms: \(-r + 50 > 0\)

Subtract 50: \(-r > -50\)

Divide by \(-1\) and FLIP SYMBOL: \(r < 50\)

Interval Notation: \((-\infty, 50)\)

Solution: r < 50

#14 20 + 16d ≥ 11(d + 20)

Distribute 11: \(20 + 16d \ge 11d + 220\)

Subtract \(11d\): \(20 + 5d \ge 220\)

Subtract 20: \(5d \ge 200\)

Divide by 5: \(d \ge 40\)

Interval Notation: \([40, \infty)\)

Solution: d ≥ 40

#15 3d - 11 ≤ 29d + 15

Method A (Variables on left):

\(3d - 29d \le 15 + 11 \implies -26d \le 26\)

Divide by \(-26\) and FLIP SYMBOL: \(d \ge -1\)

Method B (Variables on right):

\(-11 - 15 \le 29d - 3d \implies -26 \le 26d \implies -1 \le d \implies d \ge -1\)

Solution: d ≥ -1

Topic 1 Answer Key • Savvas Algebra 1 Page 2 of 3

IV

Part 4 & 5: Compound & Absolute Value Solutions with Graphs

Complete Graphs

Compound Inequalities (Probs 17-19) Full Graphical Solutions

#17 "Greater than -3 and less than 3"

-3 < x < 3

-6 -3 0 3 6

#18 8x - 11 < -19  \(\implies x < -1\)  |  10x + 14 > 14  \(\implies x > 0\)

x < -1  or  x > 0

-3 -1 0 2

#19 \(\frac{2x-1}{3} \le -7 \implies x \le -10\)  |  \(\frac{8x-2}{2} \ge 7 \implies x \ge 2\)

x ≤ -10  or  x ≥ 2

-12 -10 -4 0 2

Absolute Value Functions (Probs 20-25) Cases & Verification

#20 |x| + 5 = 3

Isolate absolute value: \(|x| = 3 - 5 \implies |x| = -2\)

Absolute value is distance: cannot equal \(-2\)!

No Solution (∅)

#21 |5x| ≥ 10

Case 1: \(5x \ge 10 \implies x \ge 2\)

Case 2: \(5x \le -10 \implies x \le -2\)

-2 2

x ≤ -2  or  x ≥ 2

#22 |d - 5| ≥ 6

Case 1: \(d - 5 \ge 6 \implies d \ge 11\)

Case 2: \(d - 5 \le -6 \implies d \le -1\)

-1 11

d ≤ -1  or  d ≥ 11

#23 |2x - 3| - 4 = 5

Isolate: \(|2x - 3| = 9\)

Case 1: \(2x - 3 = 9 \implies 2x = 12 \implies x = 6\)

Case 2: \(2x - 3 = -9 \implies 2x = -6 \implies x = -3\)

x = 6  or  x = -3

#24 |3x| < 12

Less thAND rule: \(-12 < 3x < 12\)

Divide by 3: \(-4 < x < 4\)

-4 4

-4 < x < 4

#25 |w + 2| ≤ 5

Less thAND rule: \(-5 \le w + 2 \le 5\)

Subtract 2: \(-7 \le w \le 3\)

-7 3

-7 ≤ w ≤ 3

Topic 1 Answer Key • Savvas Algebra 1 Page 3 of 3

\( 4 - |x| = 1 \)

A \( x = \pm 3 \)

B \( x = \pm 5 \)

C \( x = -5, 3 \)

D \( x = -3, 5 \)

Solution:

#11 Topic 1 (from #19)

Jamie is buying \(5\) concert tickets, and he wants his total cost to be no more than \(\$4\) above or below \(\$80\). Which inequality models the cost \(x\), in dollars, of a ticket?

A \( |5x + 4| \le 80 \)

B \( |5x - 4| \le 80 \)

C \( |80 - 5x| \le 4 \)

D \( |80 + 5x| \le 4 \)

Model:

#12 Topic 1 (from #20)

Graph the solution of the absolute value inequality:

\( 4 > |x + 1| + 2 \)

-5 -4 -3 -2 -1 0 1 2 3 4 5

Solution:

Topic 1 Assessment Review Guide • enVision Algebra 1 Page 2 of 2

Correct Choice: D (all real numbers)

Topic 1 Assessment Review Guide • Answer Key Page 1 of 2

P2

Topic 1 Assessment Review Guide — Answer Key • Part II

Problems 7 to 12

#7 Topic 1 (from #14)

Solve & Graph: \( x - (5 - 3x) \le 2x - 1 \)

1. Distribute negative: \(x - 5 + 3x \le 2x - 1\)

2. Combine: \(4x - 5 \le 2x - 1\)

3. Subtract \(2x\) and add 5: \(2x \le 4 \implies x \le 2\)

-5 -3 0 2 5

Solution: x ≤ 2 (Solid dot at 2, shade left)

#8 Topic 1 (from #16)

Write a compound inequality for the graph:

• Left ray: Open circle at \(-3\) shaded left \(\implies x < -3\)

• Right ray: Closed circle at \(2\) shaded right \(\implies x \ge 2\)

• Disjunction (two diverging rays) \(\implies\) "or"

Compound Inequality: x < -3  or  x ≥ 2

#9 Topic 1 (from #15)

Solve: \( 3x - 4 > 5 \quad\text{or}\quad 1 - 2x \ge 7 \)

• Branch 1: \(3x > 9 \implies x > 3\)

• Branch 2: \(-2x \ge 6 \implies x \le -3\) (flip symbol!)

Combined statement: \(x \le -3 \text{ or } x > 3\)

Correct Choice: A (x ≤ -3 or x > 3)

#10 Topic 1 (from #18)

Solve: \( 4 - |x| = 1 \)

1. Subtract 4: \(-|x| = 1 - 4 \implies -|x| = -3\)

2. Multiply by \(-1\): \(|x| = 3\)

3. Two cases: \(x = 3 \text{ or } x = -3 \implies x = \pm 3\)

Correct Choice: A (x = ±3)

#11 Topic 1 (from #19)

\(5\) tickets; total cost no more than \(\$4\) above or below \(\$80\).

• Total ticket cost \(= 5x\)

• Distance from \(\$80\) is at most \(\$4\): \(|5x - 80| \le 4\)

• Since \(|a - b| = |b - a|\), this is equivalent to: \(|80 - 5x| \le 4\)

Correct Choice: C (|80 - 5x| ≤ 4)

#12 Topic 1 (from #20)

Solve & Graph: \( 4 > |x + 1| + 2 \)

1. Subtract 2: \(2 > |x + 1| \implies |x + 1| < 2\)

2. "Less thAND" rule: \(-2 < x + 1 < 2\)

3. Subtract 1: \(-3 < x < 1\)

-5 -3 0 1 5

Solution: -3 < x < 1 (Open dots at -3 & 1)

Topic 1 Assessment Review Guide • Answer Key Page 2 of 2